A more general result:
[tex]1\cdot1!+2\cdot2!+\cdots+(n-1)\cdot(n-1)!+n\cdot n!=(n+1)!-1[/tex]
Why is this true?
On the left side,
[tex]1\cdot1!=(2-1)\cdot1!=2\cdot1!-1\cdot1!=2!-1![/tex]
[tex]2\cdot2!=(3-1)\cdot2!=3\cdot2!-1\cdot2!=3!-2![/tex]
[tex]3\cdot3!=(4-1)\cdot3!=4\cdot3!-1\cdot3!=4!-3![/tex]
and so on. Adding will lead to cancellation of all terms in the middle, leaving you with [tex](n+1)!-1[/tex].
Then when [tex]n=100[/tex], the value of the sum is [tex]101!-1[/tex].
